( 0,75 - 1/4 ) : (-5) + 1/15 - (1/5) : (-3)
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1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5
Giải:
a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\)
=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\)
=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\)
=\(\dfrac{7}{5}+\dfrac{-1}{4}\)
=\(\dfrac{23}{20}\)
b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\)
=\(\left[0+0\right]:\dfrac{2017}{2018}\)
=0\(:\dfrac{2017}{2018}\)
=0
c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10
a) \(0,75+\left(\dfrac{-1}{3}\right)-\dfrac{5}{18}=\dfrac{3}{4}+\left(\dfrac{-1}{3}\right)-\dfrac{5}{18}=\dfrac{5}{12}-\dfrac{5}{18}=\dfrac{5}{36}\)
c) \(\dfrac{4}{15}\cdot\dfrac{1}{3}\cdot\dfrac{15}{20}=\dfrac{4}{15}\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}=\dfrac{5}{45}\cdot\dfrac{3}{4}=\dfrac{15}{180}=\dfrac{1}{12}\)
d) \(\left(\dfrac{-1}{9}\right)\cdot\left(\dfrac{15}{22}\right):\left(\dfrac{-25}{9}\right)=\dfrac{-5}{66}:\left(\dfrac{-25}{9}\right)=\dfrac{-5}{66}\cdot\left(\dfrac{9}{-25}\right)=\dfrac{-3}{-110}=\dfrac{3}{110}\)
a) \(0,75\) + \(\dfrac{-1}{3}\) - \(\dfrac{5}{18}\)= \(\dfrac{5}{12}\) - \(\dfrac{5}{18}\) = \(\dfrac{5}{36}\)
b) \(\dfrac{4}{15}\)x \(\dfrac{1}{3}\)x \(\dfrac{15}{20}\)= 4/45 x 15/20 = 1/15
c) -1/9 x 15/22 : -25/9 = -5/66 : -25/9 = 3/110
= 4/9 : -8 - (-1/4 + 9/20) : 14/5
= 4/9 : -8 - 1/5 : 14/5
= -1/18 - 1/5 : 14/5
= -1/18 - 1/14
= -8/63
(0,75-1/4):(-5)+1/15-(1/5):(-3)
=(3/4-1/4):(-5)+1/15-(1/5)x(-1/3)
=1/2x(-1/5)+1/15-(-1/15)
=(-1/10)+1/15+1/15
=(-15/150)+10/150+1/15
=(-1/30)+1/15
=(-1/30)+2/30
=1/30